APPM 3310 1 2 3 Σ / / / / 4 / Exam 1 Write your name and your professor’s name on the front of your exam. You’re allowed one side of one sheet of letter-sized notes. You are not allowed to use textbooks, class notes, or a graphing calculator, but you may use a scientific calculator. To receive full credit on a problem you must show sufficient justification for your conclusion. 1. The following questions are independent from each other. 5 −4 (a) Let A be a 2 × 2 real matrix. Compute the solution to Ax = x1 = 1 1 solves Ax = 3 0 3 0 Solution: Since 5 −4 = , x2 = +2 1 −1 1 −2 solves Ax = if the following are true: 1 −2 , x3 = 1 2 solves Ax = 4 1 we have by the Superposition Principle x = x1 + 2x2 = 3 −1 (b) Write down a basis for the space of real symmetric 2 × 2 matrices. What is the dimension of the space? a b Solution: This is the space of all matrices of the form A = for a, b, c ∈ R. b c There are many choices for a basis for this space, but the simplest is S1 = 1 0 0 0 S2 = 0 0 0 1 S3 = 0 1 1 0 Then we can write any arbitrary symmetric 2 × 2 matrix as a b b c =a 1 0 0 0 +b 0 1 1 0 +c 0 0 0 1 Since the basis contains three elements the dimension of the space is 3. 2. The following questions are TRUE/FALSE. If your answer is TRUE you must briefly explain why it is true. If it is FALSE you must briefly explain why it is false, or give a counterexample. (a) TRUE or FALSE: If matrix A has all zeros on its main diagonal, then it is singular. Solution: The statement is false. Consider the counterexample A = 0 1 . 1 0 (b) TRUE or FALSE: For a square matrix A, if x ∈ ker A2 then x ∈ ker A Solution: The statement is false. Consider the case when A = Then, for instance, x = 1 1 ∈ ker A2 1 1 but A 0 1 0 0 , then A2 = . 0 0 0 0 6= 0 (c) TRUE or FALSE: Let A be an n × n matrix. If Ax1 = Ax2 , with x1 6= x2 , then A is not invertible. Solution: The statement is true. To see this note that ⇔ Ax1 = Ax2 A (x1 − x2 ) = 0 Since (x1 − x2 ) 6= 0 the vector is a nontrivial element of ker A so A is not invertible. (d) TRUE or FALSE: The following vectors span R2 : v1 = 2 −1 −3 , v2 = , v3 = −5 −6 4 Solution: The statement is true. Putting the vectors in a matrix and row reducing we have 2 −1 −3 4 −5 −6 ∼ 2 −1 −3 0 −3 0 The reduced matrix has two pivots so the vectors span R2 .     1 2 −1 1 3. Consider the system Ax = b where A =  −1 −2 −1  and b =  −1  −2 0 0 2 (a) Find matrices L, U and P such that P A = LU . (b) Use the result of (a) to solve the linear system for x.   1 2 −1 2 0 . Find bases for the following: (c) Now consider the slightly modified matrix B =  −1 −2 −1 −2 0 0 −2 (i) rng B (ii) corng B (iii) ker B (iv) coker B Solution: (a) To find the LU decomposition of A we do Gaussian Elimination. We have  1 2  −1 −2 −2 0  1  −1 −2 So P A = LU     −1 1 2 −1  ⇒  1R1 + R2 → R2  ⇒  −1 0 0 2R1 + R3 → R3 −2 4     2 −1 1 2 −1     0 −2 R2 ↔ R3 −2 4 −2 ⇒ ⇒ 4 −2 −1 0 −2  −1 −2  −2   ⇔       1 0 0 1 2 −1 1 0 0 1 2 −1  0 0 1   −1 −2 −1  =  −2 1 0   0 4 −2  0 1 0 −2 0 0 −1 0 1 0 0 −2 (b) We have Ax = b ⇒ ⇒ P Ax = P b  Applying our P to b we have P b =  LU x = P b.  1 2 . Then solving Ly = P b we have −1     1 0 0 1  −2 1 0  y =  2  −1 0 1 −1  ⇒    1 1 y =  2 + 2 (1) = 4  =  4  −1 + 1 (1) = 2 0 Then solving U x = y we have     1 2 −1 1  0 4 −2  x =  4  0 0 −2 0  ⇒    1 − 2 (1) + 1 (0) = −1 −1 [4 + 2 (0)] /4 = 1  =  1  = x x= 0/2 = 0 0 (c) We notice that the matrix B is exactly A with an extra column tacked on. Thus, if we did Gaussian Elimination on B we would get the same thing for the first three columns of the reduced system. We just need to apply the same row operations on the last column of B. If we do this we obtain     1 2 −1 2 1 2 −1 2  −1 −2 −1 0  ∼  0 4 −2 2  −2 0 0 −2 0 0 −2 2 (i) There are pivots in the first three columns of the reduced system. Then a basis for rng B is made up of the first three columns of the original matrix B:       1 2 −1   rng B = span  −1  ,  −2  ,  −1    −2 0 0 (ii) Since there are pivots in the first three rows of the reduced system, these rows make up a basis for corng B:  1    2 corng B = span   −1    2     0 0   4   0 ,     −2  ,  −2 2 2         (iii) To find a basis for ker B we use the reduced system to write the basic variables in terms of the one free variable x4 : x3 = (−2x4 ) / (−2) = x4 x2 = (2x3 − 2x4 ) /4 = 0 x1 = −2x2 + x3 − 2x4 = −x4 So the kernel element is given by   −1  0   x = x4   1  1 ⇒   −1       0  ker B = span   1       1 (iv) From the Fundamental Theorem of Linear Algebra we know that dim coker = m − r where r = rank A = rank AT . Since there are three pivots in the reduced version of A we have r = 3. Then, since A has m = 3 rows we have dim coker B = m − r = 3 − 3 = 0 ⇒ coker B = span {0} 4. This problem is NOT REQUIRED. Do NOT even think of attempting this problem unless you have completed problems 1-3 to your complete satisfaction. (a) Show that if A is n × n, then det (−A) = (−1)n det A (b) Prove that for odd n, any n × n skew-symmetric matrix (satisfying AT = −A) is singular. (c) Construct a nonsingular skew-symmetric matrix. Solution: (a) To change −A to A we have to multiply each row by −1. Each time we multiply a row by −1 the determinant is scaled by −1. Since we have n rows this gives det (−A) = (−1)n det A (b) We have det A = det AT = det (−A) = (−1)n det A = − det (A) since n is odd From this we see that det A = − det A ⇔ 2 det A = 0 ⇔ det A = 0 ⇔ A is singular (c) We know that we need n even for skew-symmetric A to be nonsingular. The simplest example is A= 0 1 −1 0